poj2965 bfs+位运算 异或真他妈的NB

#include
#include
using namespace std;
int maxn = (1 << 16) + 10, num[20] = { 0 };
int queue[65540] = { 0 }, visit[65540] = { 0 }, father[65540] = { 0 }, way[65540] = { 0 }, number[655540] = { 0 };
void innit() //初始化16种情况
{
	int i = 0,row=0,colum=0,m=0,n=0;
	for (i = 0; i < 16; i++)
	{
		row = i / 4, colum = i % 4;
		for (m = 0; m < 4; m++)
		{
			if (m != row)
				num[i] += 1 << (4 * m + colum);
		}
		for (n = 0; n < 4; n++)
		{
			if (n != colum)
				num[i] += 1 << (4 * row + n);
		}
		num[i] += 1 << (4 * row + colum);
	}
}
void bfs(int p)
{
	int front = 0, i = 0, tail = 0;
	queue[front] = p;
	while (1)
	{
		if (!queue[front]) break;
		for (i = 0; i < 16; i++)
		{
			if (visit[queue[front] ^ num[i]]) continue;//记忆化剪枝
			tail++;
			queue[tail] = queue[front] ^ num[i];//异或翻新
			father[tail] = front;// 记录父节点和子节点之间的联系
			number[tail] = number[front]+1;
			way[tail] = i;//记录每一节点的翻转位置
			visit[queue[front] ^ num[i]] = 1;
		}
		front++;
	}
	cout << number[front] << endl;
	for (i = front; i; i = father[i])
	{
		cout << way[i] / 4 + 1 << " " << way[i] % 4 + 1<> sign;
			if(sign=='+') s += 1 << (4 * i + j);
		}
		getchar();
	}
	bfs(s);
	return 0;
}